A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is `5ms^(-1)` and the speed is increasing at a rate of `2ms^(-2)`. At this instant, the magnitude of the net acceleration will be

Options

A. `3.2ms^(-2)`
B. `2ms^(-2)`
C. `2.5ms^(-2)`
D. `4.3ms^(-2)`

Answer

A

Explanation

Magnitude of the net acceleration `a=sqrt(a_t^2+a_n^2)`
where, `a_t "rate of change of speed "=2 ms ^(-2)`
`a_n=(v^2)/(R)-(5)^2/(10)=2.5rArr a=sqrt((2)^2+(2.5)^2)=3.2ms^(-2)`